Let $ X, Y $ be smooth affine varieties over $ \mathbb C $. Let $ T : X \rightarrow Y $ be a dominant quasi-finite morphism and let $ T^\# : \mathbb C[Y] \rightarrow \mathbb C[X] $ be the resulting map on coordinate rings.

($ T $ being quasi-finite means that it can be factored as an open embedding followed by a finite morphism. It also means that $T^\# $ is injective and every element of $\mathbb C[X] $ is algebraic over $ \mathbb C[Y] $.)

Let $ D(X) $ denote the ring of differential operators on $ X $. Is the following statement is true?

Let $ d \in D(X) $. Suppose that $ d(T^\#(f)) = 0 $ for all $ f \in \mathbb C[Y] $. Then $ d = 0 $.

  • 1
    $\begingroup$ I guess you want to add the condition that $T$ is dominant for $T^\#$ to be injective (otherwise a closed embedding is an example of a quasi-finite morphism). $\endgroup$ Commented Apr 13, 2018 at 17:26
  • $\begingroup$ I agree and made the edit. $\endgroup$ Commented Apr 13, 2018 at 20:00

1 Answer 1


Edit: I think skipped a step in the original argument - it is not immediately clear to me that $d$ commutes with $\mathbb C[Y]$ in $D(X)$ unless $d$ is a derivation. I have added an inductive argument for this below.

I believe this is true. Here is a sketch of an argument.

We will prove this by induction on the degree of the differential operator $d$. It is certainly true for differential operators of degree 0.

Suppose $d\in D(X)$ is of degree $\leq k$, and $d(f)=0$ for all $f\in \mathbb C[Y]$ (which I identify with a subring of $\mathbb C[X]$).

Lemma: We have $[d,f]=0$ for all $f\in \mathbb C[Y]$.

Proof of lemma: The differential operator $[d,f]$ is of order $\leq k-1$, and $[d,f](g)=0$ for all $g\in \mathbb C[Y]$. The lemma now follows from the inductive hypothesis.

It remains to show that any $\mathbb C[Y]$-linear differential operator $d\in D(X)$ is zero.

Consider the finite field extension $$K := \mathbb{C}(Y) \subseteq L := \mathbb{C}(X)$$

Note that $D(X)$ injects in to $D(L/\mathbb C)$ and $d$ lands in the subring $D(L/K)$. In other words $d$ may be considered as an element of the ring $D(L/K)$ (here I am using Grothendieck's definition of differential operators for a $K$-algebra).

I claim that $D(L/K)=L$, and thus $d=0$. Indeed, $L$ is a smooth $K$-algebra of dimension $0$ (as we are in characteristic $0$), and thus:

  1. The ring of $D(L/K)$ is generated by derivations $Der_K(L,L)$ and $L$.
  2. $Der_K(L,L)=0$.
  • $\begingroup$ This looks good to me. Thanks! Can you provide a reference for the last two facts? $\endgroup$ Commented Apr 13, 2018 at 20:18
  • 1
    $\begingroup$ According to a comment by Mariano Suarez-Alvarez in an answer to this question: math.stackexchange.com/questions/1974919/… , there is a proof of Fact 1 in the last chapter of McConner and Robson's book Noncommutative Noetherian Rings. However, I can't seem to get access to a copy at the moment to check. $\endgroup$ Commented Apr 13, 2018 at 20:53
  • 2
    $\begingroup$ Fact 2 is much more elementary. It is equivalent to the assertion that $\Omega^1_{L/K}=0$, which in turn is equivalent to the fact that $L/K$ is separable (immediate in characteristic $0$). The wikipedia page en.wikipedia.org/wiki/… indicates that there is a reference in Milne's book on etale cohomology, for example. $\endgroup$ Commented Apr 13, 2018 at 20:55
  • 1
    $\begingroup$ Thanks! Yes, I think that I know how to prove Fact 2. $\endgroup$ Commented Apr 13, 2018 at 22:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.